# B Definition of the limit of a sequence

1. Jul 27, 2016

### Math_QED

The definition of a limit of a sequence,
if the limit is finite, is:

lim n >infinity un (un is a sequence) = l

<=>

∀ε> 0, ∃N: n > N => |un - l| < ε

This just means that un for n > N has to be a number for which: l -ε < un < l + ε

Now, I'm wondering, can't we just say:

n > N => |un -l| < kε with k a real number larger than zero?

Thanks in advance

2. Jul 27, 2016

### Staff: Mentor

What for? There is no mathematical difference between $ε$ and $kε$ for a constant $k$. Why do you want to complicate something without need that often is hard to understand to newbies anyway.

3. Jul 28, 2016

### micromass

Staff Emeritus
I disagree that it is without need. You often see in mathematical derivations that something is $<3\varepsilon$ and that that's good enough. Such proofs are even known as $3\varepsilon$-proofs. I do agree that what's in the OP shouldn't be the standard definition as it already is hard enough. But it should be made much more clearer in analysis books that $<k\varepsilon$ is good enough.

4. Jul 28, 2016

### PeroK

When people are learning limits often a lot of complexity comes about for the case where $\epsilon$ is not small. Also, sometimes it complicates things to get the $< \epsilon$ in the definition and it's simpler to get $\le \epsilon$. I always thought a better (equivalent) definition would be.

$\exists \epsilon_0 > 0: \forall \epsilon \in (0, \epsilon_0), \exists N: n > N \Rightarrow |u_n - L| \le \epsilon$

In fact, as I don't have to sit exams, this is the one I use!

You could, if you wanted to, add your $k$:

$\exists k, \ \epsilon_0 > 0: \forall \epsilon \in (0, \epsilon_0), \exists N: n > N \Rightarrow |u_n - L| \le k\epsilon$

5. Jul 28, 2016

### Staff: Mentor

I learned the notation $\forall \epsilon \exists N(\epsilon)$ or $\forall \epsilon \exists N_\epsilon$ so there has never been a doubt that a the find of $N$ depends on the choice of $\epsilon$ and a constant factor can easily be transformed into $N_\epsilon$.

I find an additional "variable" in the definition would only cause more difficulties, such as to think the small sized $\epsilon$ could be blown up beyond any natural number. Although still correct, it's not helpful. (IMO, but I don't want hell to break lose ...)

6. Jul 28, 2016

### Staff: Mentor

I agree, however, wouldn't that simply transform the difficulties to later concepts, because it then makes the difference between an open and a closed neighborhood. I always thought this to be the reason for the strict < . Otherwise one has to explain why all of a sudden we change from closed to open.

7. Jul 28, 2016

### micromass

Staff Emeritus
You're right. But when I first learned analysis at university, one of the very first exercises we did was to show that we could replace all notions of $<$ with $\leq$. I remember that I found it somewhat difficult to grasp, and I didn't really see its importance. In retrospect, I'm happy that we did this.

8. Jul 28, 2016

### Math_QED

Yes, this is exactly why I made this thread. In the following proof, we encounter something similar:

Theorem: If limn-> ∞un = l1 and limn-> ∞vn = l2

limn-> ∞(un + vn) = l1 + l2

Proof: We can say for all ε>0

∃N: n> N => |un-l1|<ε
∃N': n> N' => |vn-l2|<ε

Now, we observe, that:

|un + vn - (l1 + l2)| = |(un - l1) + (vn - l2)| =< |un - l1| + |vn - l2|

So, for n > max{N, N'}, we find:

|un + vn - (l1 + l2)| = |(un - l1) + (vn - l2)| =< |un - l1| + |vn - l2| < ε + ε

=> |un + vn - (l1 + l2)| < 2ε

Now, the book does something I'm unsure about:

They say, choose ε' > 0, and let ε = ε'/2. Then we have, for all ε' > 0

n > max {N, N'} => |un + vn - (l1 + l2)| < ε' and what we wanted to proof follows.

It makes sense to me that |un + vn - (l1 + l2)| < 2ε would be enough to proof, as Micromass already said (that's why I created this thread), but I do not fully understand why we can say that ε = ε'/2? Is this, because ε>0 => ε'/2 > 0?

Probably this is a dumb question, but I'm self studying this to prepare for university mathematics to have at least a bit experience with some more abstract mathematics. I got this analysis book from my brother who studies physics.

Last edited: Jul 28, 2016
9. Jul 28, 2016

### Staff: Mentor

It will get better with time. Wait until you will finish your proofs as soon as some $\epsilon$ show up
and all the work now, like $\epsilon' = 2 \epsilon$, $\epsilon' = \frac{1}{3} \epsilon$ or $\epsilon = \min\{\epsilon_1, \epsilon_2\}$ will be classified under the label: "boring technical issues". Unfortunately until then, the boring stuff has to be learned, too. (see post #7)

10. Jul 28, 2016

### Math_QED

Well I do not think it's boring at all ;)

11. Jul 28, 2016

### PeroK

The other aspect of this is the powerful logical thinking required. For example, the negation of the limit. This confounds a lot of people when they first meet it. How do you say that a sequence does not have a limit of $L$? This logical thinking is vital, even if the epsilons and deltas have been replaced by other more abstract concepts.

12. Jul 28, 2016

### Math_QED

I'm not quite sure about this one:

Would this be:

∃ε > 0, ∀N: n>N => |un - L | >= ε

or something like..

∀ε > 0, there is no N: n>N => |un - L | < ε

EDIT: could someone give a look at post #8?

13. Jul 28, 2016

### PeroK

Neither, I'm afraid. Have a good think!

14. Jul 28, 2016

### micromass

Staff Emeritus
Both are incorrect.

15. Jul 28, 2016

### Svein

Not only newbies. In the University I once had a very old-fashioned professor. When proving a limit, he chose an ε and a δ, and after about half an hour he ended up with ".. less than 2ε". Of course, we were satisfied, but he started the next lecture by choosing a δ half as big as he had done in the previous lecture and did the whole proof over!

On another note, I remember a proof in multidimensional complex analysis which ended up with "... less that 10 000ε, which is small when ε is small."

16. Jul 28, 2016

### Math_QED

∃ε > 0, ∀N: n>N ∧ |un - L | >= ε

or: ∃ε > 0, ∀N: |un - L | >= ε => n=<N

I used neg(a=>b) <=> a Λ neg(b) for the first and contraposition for the second. I had to look up some logic first because this is something we were not taught in high school.

Can someone look at post #8, this is where I am stuck in the proof.

17. Jul 28, 2016

### PeroK

It's not a dumb question. The crux of the matter is that if you choose any $\epsilon > 0$, then $\epsilon ' = \epsilon/2 > 0$. And, for this $\epsilon '$ you can find $N_1, N_2$ such that:

$n > N_1 \Rightarrow |u_n - l_1| < \epsilon ' = \epsilon/2$

And

$n > N_2 \Rightarrow |v_n - l_2| < \epsilon ' = \epsilon/2$

Now, $n > max\{N_1, N_2 \}$ gives you what you want.

Does that make sense?

Personally, I would do this without introducing $\epsilon '$. I'd just say that because $\epsilon/2 > 0$ we can find an $N_1$ such that:

$n > N_1 \Rightarrow |u_n - l_1| < \epsilon/2$ etc.

PS I can see the book did things a bit differently. I definitely don't like using $N'$ for the $l_2$ limit. Using $N_1, N_2$ seems much more logical. Especially as $N'$ has nothing to do with $\epsilon '$

Last edited: Jul 28, 2016
18. Jul 28, 2016

### Math_QED

In fact, the book used N1 and N2 but I wrote down the proof without the book. You are right that this is more logical. I believe your explanation helped me further. Thank you for that! I will think about it some more. In the meantime, what do you think about the negation of the limit?

19. Jul 28, 2016

### PeroK

For the negation, think about the sequence:

$1, 1/2, 1, 1/4, 1, 1/8, \dots$

And why it does not converge to either $0$ or $1$ (or anything else!).

20. Jul 28, 2016

### Math_QED

Is what I wrote in post #16 wrong?

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